notes/OJ notes/pages/Leetcode Search-A-2D-Matrix-II.md

Leetcode Search-A-2D-Matrix-II

2022-09-03 14:57

Algorithms:

#algorithm #divide_and_conquer

Data structures:

#DS #array

Difficulty:

#coding_problem #difficulty_medium

Additional tags:

#leetcode #CS_list_need_practicing

Revisions:

N/A

Links:

• Link to problem

---

Problem

Write an efficient algorithm that searches for a value target in an m x n integer matrix matrix. This matrix has the following properties:

• Integers in each row are sorted in ascending from left to right.
• Integers in each column are sorted in ascending from top to bottom.

Examples

Example 1:

**Input:** matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 5
**Output:** true

Example 2:

**Input:** matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 20
**Output:** false

Constraints

• m == matrix.length
• n == matrix[i].length
• 1 <= n, m <= 300
• -109 <= matrix[i][j] <= 109
• All the integers in each row are sorted in ascending order.
• All the integers in each column are sorted in ascending order.
• -109 <= target <= 109

Thoughts

[!summary] This is a #divide_and_conquer problem.

It's divide and conquer, because every time we do a action, the problem is smaller.

Start from the top-right, (alternatively, bottom-left), because walking left makes the number smaller, and down makes the number bigger.

Solution

class Solution {
public:
  bool searchMatrix(vector<vector<int>> &matrix, int target) {
    // search from top-right
    int c = matrix[0].size() - 1;
    int r = 0;
    int m = matrix.size();
    while (c >= 0 && r < m) {
      if (matrix[r][c] > target) {
        c--;
      } else if (matrix[r][c] < target) {
        r++;
      } else {
        return true;
      }
    }

    return false;
  }
};